Question

A single slit diffraction pattern is formed with white light. For what wavelength of light the 4th secondary maximum in diffraction pattern coincides with the 3rd secondary maximum in the pattern of light of wavelength '$\lambda$'?

• A. $\frac{5\lambda}{7}$
• B. $\frac{7\lambda}{9}$
• C. $\frac{3\lambda}{4}$
• D. $\frac{9\lambda}{13}$

Hint:

Don't confuse Interference (YDSE) with Diffraction!
In YDSE, Maxima are at $n\lambda$.
In Single Slit, Maxima are at $(n + 0.5)\lambda$ and Minima are at $n\lambda$. This half-integer shift is crucial!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In single slit diffraction, the position of the $n$-th secondary maximum is given by $y_n = (2n + 1)\frac{\lambda D}{2d}$.

Step 2: Formula Application:
For wavelength $\lambda'$, $n = 4$: $y_4 = (2 \times 4 + 1)\frac{\lambda' D}{2d} = \frac{9 \lambda' D}{2d}$. For wavelength $\lambda$, $n = 3$: $y_3 = (2 \times 3 + 1)\frac{\lambda D}{2d} = \frac{7 \lambda D}{2d}$.

Step 3: Explanation:
Set $y_4 = y_3$: $\frac{9 \lambda' D}{2d} = \frac{7 \lambda D}{2d}$ $9 \lambda' = 7 \lambda \implies \lambda' = \frac{7\lambda}{9}$.

Step 4: Final Answer:
The required wavelength is $\frac{7\lambda}{9}$.